1. Statement of the problem: Solve $ (i - x)^3 = -1$. 2. Formula and rule: For complex numbers, to solve $z^3 = w$ we write $w$ in polar form and take cube roots using $$z = e^{i(\arg(w)+2\pi k)/3}\,,\quad k=0,1,2$$. Note that $\arg(w)$ denotes a chosen argument and different values of $k$ give the three distinct roots. 3. Write -1 in polar form: $$-1 = e^{i(\pi + 2\pi n)}\,,\quad n\in\mathbb{Z}$$. 4. Take cube roots: $$i - x = e^{i(\pi + 2\pi n)/3}\,,\quad n=0,1,2$$. 5. Evaluate $n=0$: $$i - x = e^{i\pi/3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$. Hence $$x = i - \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -\frac{1}{2} + i\left(1 - \frac{\sqrt{3}}{2}\right) = -\frac{1}{2} + i\frac{2-\sqrt{3}}{2}$$. 6. Evaluate $n=1$: $$i - x = e^{i\pi} = -1$$. Hence $$x = i - (-1) = 1 + i$$. 7. Evaluate $n=2$: $$i - x = e^{i5\pi/3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$$. Hence $$x = i - \left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -\frac{1}{2} + i\left(1 + \frac{\sqrt{3}}{2}\right) = -\frac{1}{2} + i\frac{2+\sqrt{3}}{2}$$. 8. Final answer: The three solutions are $$x = -\frac{1}{2} + i\frac{2-\sqrt{3}}{2},\quad x = 1 + i,\quad x = -\frac{1}{2} + i\frac{2+\sqrt{3}}{2}$$.